$\begin{aligned} &f(n)=(2n-3)(n+1) \\\\ &h(t)=-8(3-t) \end{aligned}$ $(h\circ f) (1.5)=$
Solution: Let's start by rewriting $(h\circ f) (1.5)$ as $h(f(1.5))$. When evaluating composite functions, we work our way inside out. To evaluate $h(f(1.5))$, let's first evaluate $f(1.5)$. Then we'll plug that result into $h$ to find our answer. Let's evaluate $f({1.5})$. $\begin{aligned}f(n)&=(2n-3)(n+1)\\\\ f({1.5})&=(2({1.5})-3)({1.5}+1)~~~~~~~~~~\text{Plug in }n={1.5}\\\\ &=(0)(2.5)\\\\ &={0}\end{aligned}$ We now know that $h(f({1.5}))$ is the same as $h({0})$ because $f({1.5}) = {0}$. Let's evaluate $h({0})$. $\begin{aligned}h(t)&=-8(3-t)\\\\ h({{0}})&=-8(3-{0})~~~~~~~~~~\text{Plug in }t={0}\\\\ &=(-8)(3)\\\\ &=-24\\\\\end{aligned}$ The answer: $(h\circ f)(1.5) =-24$